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A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net force acts on the otter along the incline?

Answer :

Answer:

The net force acting on the otter along the incline is 13.96 N.

Explanation:

It is given that,

Mass of the otter, m = 2 kg

Distance covered by otter, d = 85 cm = 0.85 m

It takes 0.5 seconds.

We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

Here, u = 0 (at rest)

[tex]d=\dfrac{1}{2}at^2[/tex]

[tex]a=\dfrac{2d}{t^2}[/tex]

[tex]a=\dfrac{2\times 0.85}{0.5^2}[/tex]

[tex]a=6.8\ m/s^2[/tex]

The net force acting on the otter along the incline is given by :

F = ma

[tex]F=2\ kg\times 6.8\ m/s^2[/tex]

F = 13.6 N

So, the net force acting on the otter along the incline is 13.96 N. Hence, this is the required solution.

Answer:

[tex]F=13.6\rm N[/tex] Force acts on the otter along the incline

Explanation:

Given information:

Incline length [tex]s=\rm 85cm=0.85m[/tex]

Time [tex]t=0.5\rm sec[/tex]

Mass of otter [tex]m=2\rm kg[/tex],

Initial velocity [tex]u=0[/tex] as otter is in rest

Muddy incline so we can assume surface as friction less

By use equation of motion,

[tex]s=ut+\frac{1}{2} a t^2[/tex]

[tex]s=ut+\frac{1}{2} a t^2=0\times t+\frac{1}{2}\times a\times t^2[/tex]

[tex]0.85=0+\frac{1}{2} a (0.5)^2\\\\a=\frac{2\times0.85}{0.5^2}=6.8\rm m/s^2[/tex]

The net force act on the otter is

[tex]F=ma[/tex]

[tex]F=2\rm kg\times6.8\rm m/s^2[/tex]

[tex]F=13.6\rm N[/tex]

Hence [tex]F=13.6\rm N[/tex] force acts on the otter along the incline

For more details please refer link:

https://brainly.com/question/13704950?referrer=searchResults

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